Evaluate $~~\int^2_0 xe^{4x}dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac4{9}e^6-\dfrac1{9}$ (Choice B) B $\dfrac4{9}e^6+\dfrac1{9}$ (Choice C) C $\dfrac9{16}e^8-\dfrac1{16}$ (Choice D) D $\dfrac7{16}e^8+\dfrac1{16}$
Explanation: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=e^{4x} dx\,$. Then $~du = dx~$ and $~v = \int e^{4x}dx = \dfrac14e^{4x}\,$. Integration by parts gives $ \int^2_0 xe^{4x}dx = x\cdot\dfrac14e^{4x}\Bigg]_0^2-\int^2_0\dfrac14e^{4x}dx$ $ ~\,=\dfrac x4e^{4x}-\dfrac1{16}e^{4x}\Bigg]^2_0 $ $ ~\,= \bigg(\dfrac24e^8-\dfrac1{16}e^8\bigg)-\bigg(0-\dfrac1{16}\bigg)$ $ ~\,=\dfrac7{16}e^8+\dfrac1{16}\,$.